Question: Solve for $d$. $\dfrac32=5d-\dfrac12 $
Explanation: Let's add and then divide to get $d$ by itself. $\begin{aligned} \dfrac32&=5d-\dfrac12 \\ \\ \dfrac32{+\dfrac12} &=5d-\dfrac12 {+\dfrac12}~~~~~~{\text{add }\dfrac12} \text{ to each side}\\ \\ \dfrac32{+\dfrac12}&=5d-\cancel{ \dfrac12} {{+}\cancel{{\dfrac12}}}\\ \\ \dfrac32{+\dfrac12}&=5d \end{aligned}$ $\begin{aligned}2&= 5d \\ \\ \dfrac{2}{{5}} &= \dfrac{5d}{{5}} ~~~~~~~\text{divide each side by } {5} \text{ to get } d \text{ by itself }\\ \\ \dfrac{2}{{5}} &= \dfrac{\cancel{5}d}{\cancel{{5}}} \\ \\ \dfrac{2}{{5}}&= d \end{aligned}$ The answer: $d={\dfrac25}$ Let's check to make sure. $\begin{aligned} \dfrac32&=5d-\dfrac12 \\\\ \dfrac32&\stackrel{?}{=} 5\left({\dfrac25}\right)-\dfrac12 \\\\ \dfrac32&\stackrel{?}{=} 2-\dfrac12 \\\\ \dfrac32 &= \dfrac32 ~~~~~~~~~~\text{Yes!} \end{aligned}$